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JEE Main 2021 (22 Jul, Shift 2) · Prism, Dispersion & Scattering

Prism, Dispersion & Scattering question from JEE Main 2021 (22 Jul, Shift 2)

Q63JEE Main 2021 · 22 JUL · SHIFT 2Medium

Prism, Dispersion & Scattering

A ray of light passing through a prism (μ=3\mu = \sqrt{3}) suffers minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. Then, the angle of prism is _____________ (in degrees).

Numerical answer type. Work it out, then open the answer.

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Answer: 60

  1. At minimum deviation r=A2r = \frac{A}{2}, and here i=2ri = 2r.
  2. sin⁡2r=3sin⁡r⇒2sin⁡rcos⁡r=3sin⁡r⇒cos⁡r=32\sin 2r = \sqrt{3}\sin r \Rightarrow 2\sin r\cos r = \sqrt{3}\sin r \Rightarrow \cos r = \frac{\sqrt{3}}{2}
  3. r=30∘⇒A=2r=60∘r = 30^\circ \Rightarrow A = 2r = 60^\circ, so the answer is 60.
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